Question: In the diagram, $PQR$ is a straight line.  What is the value of $x$?

[asy]
draw((-2,0)--(8,0),linewidth(0.7)); draw((8,0)--(5,-5.5)--(0,0),linewidth(0.7));
label("$P$",(-2,0),W); label("$Q$",(0,0),N); label("$R$",(8,0),E); label("$S$",(5,-5.5),S);

label("$136^\circ$",(0,0),SW); label("$64^\circ$",(7,0),S);

label("$x^\circ$",(5,-4.5));

[/asy]
Since $\angle PQS$ is an exterior angle of $\triangle QRS$, then $\angle PQS=\angle QRS+\angle QSR$, so $136^\circ = x^\circ + 64^\circ$ or $x = 136-64=\boxed{72}$.